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\section{Sylvester 阵与结式}

本节, 我们用行列式研究整式的一些性质.

为方便, 我们作一个有名的 (有名字的) 假定.
我们会反复地说它.

\begin{assumption}\label{standing-assumption}
    设 \(n\), \(m\) 是非负整数,
    且 \(m + n \geq 1\).
    设
    \begin{align*}
        f(x) &
        = \sum_{k = 0}^{n} {a_k x^{n-k}}
        = a_0 x^n + a_1 x^{n-1} + \dots + a_n,
        \\
        g(x) &
        = \sum_{k = 0}^{m} {b_k x^{m-k}}
        = b_0 x^m + b_1 x^{m-1} + \dots + b_m.
    \end{align*}
    为方便, 我们约定:
    若 \(k < 0\) 或 \(k > n\), 则 \(a_k = 0\);
    若 \(k < 0\) 或 \(k > m\), 则 \(b_k = 0\).
\end{assumption}

\begin{definition}
    设假定~\ref{standing-assumption} 成立.
    定义 \(f\), \(g\) 的
    \emph{Sylvester} \pinjino{sirvisite}
    \emph{阵}%
    为 \(m + n\)~级阵
    \(S(f, g; n, m)\),
    其中,
    \begin{align*}
        [S(f, g; n, m)]_{i,j}
        = \begin{cases}
              a_{i-j},     & j \leq m; \\
              b_{i-(j-m)}, & j > m.
          \end{cases}
    \end{align*}
    \(f\), \(g\) 的 Sylvester 阵的行列式
    \(\det {(S(f, g; n, m))}\)
    是 \(f\), \(g\) 的\emph{结式}.
\end{definition}

我们看几个例.

\begin{example}
    设假定~\ref{standing-assumption} 成立.
    设 \(n = 2\), 且 \(m = 5\).
    则 \(f\), \(g\) 的 Sylvester 阵
    \(S(f, g; 2, 5)\) 是
    \begin{align*}
        \begin{bmatrix}
            a_{0} & a_{-1} & a_{-2} & a_{-3} & a_{-4} & b_{0} & b_{-1} \\
            a_{1} & a_{0}  & a_{-1} & a_{-2} & a_{-3} & b_{1} & b_{0}  \\
            a_{2} & a_{1}  & a_{0}  & a_{-1} & a_{-2} & b_{2} & b_{1}  \\
            a_{3} & a_{2}  & a_{1}  & a_{0}  & a_{-1} & b_{3} & b_{2}  \\
            a_{4} & a_{3}  & a_{2}  & a_{1}  & a_{0}  & b_{4} & b_{3}  \\
            a_{5} & a_{4}  & a_{3}  & a_{2}  & a_{1}  & b_{5} & b_{4}  \\
            a_{6} & a_{5}  & a_{4}  & a_{3}  & a_{2}  & b_{6} & b_{5}  \\
        \end{bmatrix}
        =
        \begin{bmatrix}
            a_0 & 0   & 0   & 0   & 0   & b_0 & 0   \\
            a_1 & a_0 & 0   & 0   & 0   & b_1 & b_0 \\
            a_2 & a_1 & a_0 & 0   & 0   & b_2 & b_1 \\
            0   & a_2 & a_1 & a_0 & 0   & b_3 & b_2 \\
            0   & 0   & a_2 & a_1 & a_0 & b_4 & b_3 \\
            0   & 0   & 0   & a_2 & a_1 & b_5 & b_4 \\
            0   & 0   & 0   & 0   & a_2 & 0   & b_5 \\
        \end{bmatrix}.
    \end{align*}
\end{example}

\begin{example}
    设假定~\ref{standing-assumption} 成立.
    设 \(m = n = 1\).
    则 \(f(x) = a_0 x + a_1\),
    且 \(g(x) = b_0 x + b_1\).
    则 \(f\), \(g\) 的 Sylvester 阵
    \begin{align*}
        S(f, g; 1, 1) = \begin{bmatrix}
                            a_0 & b_0 \\
                            a_1 & b_1 \\
                        \end{bmatrix},
    \end{align*}
    且
    \begin{align*}
        \det {(S(f, g; 1, 1))} = a_0 b_1 - b_0 a_1.
    \end{align*}
\end{example}

\begin{example}\label{emp:morniszero}
    设假定~\ref{standing-assumption} 成立.
    设 \(n = 0\).
    则 \(f(x) = a_0\).
    则 \(m\)~级阵 \(S(f, g; 0, m)\) 适合,
    \begin{align*}
        [S(f, g; 0, m)]_{i,j} = a_{i-j}
        = \begin{cases}
              a_0, & i = j;       \\
              0,   & \text{别的情形}.
          \end{cases}
    \end{align*}
    则 \(S(f, g; 0, m) = a_0 I_m\),
    且 \(\det {(S(f, g; 0, m))} = a_0^m\).

    类似地, 若 \(m = 0\),
    则 \(n\)~级阵 \(S(f, g; n, 0) = b_0 I_n\),
    且 \(\det {(S(f, g; n, 0))} = b_0^n\).
\end{example}

\begin{example}
    设假定~\ref{standing-assumption} 成立.
    设 \(a_0 = 0 = b_0\).
    则 \(k \leq 0\) 时, \(a_k = 0 = b_k\).
    注意, 当 \(j \leq m\) 时,
    \([S(f, g; n, m)]_{1,j} = a_{1-j}\),
    而 \(1 - j \leq 0\),
    则 \([S(f, g; n, m)]_{1,j} = 0\).
    当 \(j > m\) 时,
    \([S(f, g; n, m)]_{1,j} = b_{1 - (j - m)}\),
    而 \(1 - (j - m) \leq 0\),
    则 \([S(f, g; n, m)]_{1,j} = 0\).
    则 \(S(f, g; n, m)\) 的行~\(1\) 的元全是 \(0\).
    则 \(\det {(S(f, g; n, m))} = 0\).
\end{example}

我们给一个研究 Sylvester 阵的可能的原因.
设假定~\ref{standing-assumption} 成立.
设数 \(c\) 适合, \(f(c) = 0 = g(c)\).
则
\begin{align*}
    a_0 c^n + a_1 c^{n-1} + \dots + a_n & = 0, \\
    b_0 c^m + b_1 c^{m-1} + \dots + b_m & = 0.
\end{align*}
若我们分别以 \(c^{m-1}\), \(c^{m-2}\), \(\dots\), \(1\) 乘等式~1,
且分别以 \(c^{n-1}\), \(c^{n-2}\), \(\dots\), \(1\) 乘等式~2,
我们有
\begin{align*}
    a_0 c^{m+n-1} + a_1 c^{m+n-2} + \dots + a_{n-1} c^{m} + a_n c^{m-1}   & = 0,   \\
    a_0 c^{m+n-2} + a_1 c^{m+n-3} + \dots + a_{n-1} c^{m-1} + a_n c^{m-2} & = 0,   \\
                                                                          & \dots, \\
    a_0 c^{n} + a_1 c^{n-1} + \dots + a_{n-1} c + a_n 1                   & = 0,   \\
    b_0 c^{m+n-1} + b_1 c^{m+n-2} + \dots + b_{m-1} c^{n} + b_m c^{n-1}   & = 0,   \\
    b_0 c^{m+n-2} + b_1 c^{m+n-3} + \dots + b_{m-1} c^{n-1} + b_m c^{n-2} & = 0,   \\
                                                                          & \dots, \\
    b_0 c^{m} + b_1 c^{m-1} + \dots + b_{m-1} c + b_m 1                   & = 0.
\end{align*}
回想, 当 \(k < 0\) 或 \(k > n\) 时, \(a_k = 0\),
且当 \(k < 0\) 或 \(k > m\) 时, \(b_k = 0\).
于是, 我们可写
\begin{align*}
    a_{0} c^{m+n-1} + a_{1} c^{m+n-2} + \dots + a_{m+n-2} c + a_{m+n-1} 1  & = 0,   \\
    a_{-1} c^{m+n-1} + a_{0} c^{m+n-2} + \dots + a_{m+n-3} c + a_{m+n-2} 1 & = 0,   \\
                                                                           & \dots, \\
    a_{1-m} c^{m+n-1} + a_{2-m} c^{m+n-2} + \dots + a_{n-1} c + a_{n} 1    & = 0,   \\
    b_{0} c^{m+n-1} + b_{1} c^{m+n-2} + \dots + b_{m+n-2} c + b_{m+n-1} 1  & = 0,   \\
    b_{-1} c^{m+n-1} + b_{0} c^{m+n-2} + \dots + b_{m+n-3} c + b_{m+n-2} 1 & = 0,   \\
                                                                           & \dots, \\
    b_{1-n} c^{m+n-1} + b_{2-n} c^{m+n-2} + \dots + b_{m-1} c + b_{m} 1    & = 0.
\end{align*}
作 \(1 \times (m + n)\)~阵
\(C = [c^{m+n-1}, c^{m+n-2}, \dots, c, 1]\).
则
\begin{align*}
    S(f, g; n, m)^{\mathrm{T}}\, C^{\mathrm{T}} = 0.
\end{align*}
注意, 线性方程组
\(S(f, g; n, m)^{\mathrm{T}}\, X = 0\)
有非零的解 \(X = C^{\mathrm{T}}\).
故, 由 Cramer 公式 (见第~1~章, 节~\sekcio{21})
与反证法,
\(S(f, g; n, m)^{\mathrm{T}}\) 的行列式一定是 \(0\).
故 \(\det {(S(f, g; n, m))} = 0\).

\begin{theorem}\label{thm:necessary-condition-for-the-existence-of-a-common-root}
    设假定~\ref{standing-assumption} 成立.
    设存在数 \(c\), 使 \(f(c) = 0 = g(c)\).
    则
    \begin{align*}
        \det {(S(f, g; n, m))} = 0.
    \end{align*}
\end{theorem}

我们进一步地研究 Sylvester 阵.

\begin{theorem}
    设假定~\ref{standing-assumption} 成立.
    作 \(1 \times (m+n)\)~阵 \(C_{m+n}\),
    其中, \([C_{m+n}]_{1,j} = x^{m+n-j}\).
    设 \(Y\) 是 \((m+n) \times 1\)~阵.
    记
    \begin{align*}
        u(x) = \sum_{k = 0}^{m-1} {[Y]_{k+1,1} x^{m-1-k}},
        \quad
        v(x) = \sum_{k = 0}^{n-1} {[Y]_{m+1+k,1} x^{n-1-k}}.
    \end{align*}

    (1)
    \(C_{m+n}\, S(f, g; n, m)\) 是 \(1 \times (m + n)\)~阵,
    且
    \begin{align*}
        [C_{m+n}\, S(f, g; n, m)]_{1,j}
        = \begin{cases}
              f(x)\, x^{m-j},     & j \leq m; \\
              g(x)\, x^{n-(j-m)}, & j > m.
          \end{cases}
    \end{align*}

    (2)
    \(C_{m+n}\, S(f, g; n, m)\, Y\) 是 \(1\)~级阵, 且
    \begin{align*}
        [C_{m+n}\, S(f, g; n, m)\, Y]_{1,1} = f(x)\, u(x) + g(x)\, v(x).
    \end{align*}

    (3)
    记
    \begin{align*}
        f(x)\, u(x) + g(x)\, v(x) = \sum_{k = 0}^{m+n-1} {c_k x^{m+n-1-k}}.
    \end{align*}
    则 \(S(f, g; n, m)\, Y\) 是 \((m+n) \times 1\)~阵, 且
    \begin{align*}
        c_k = [S(f, g; n, m)\, Y]_{k+1,1}.
    \end{align*}
\end{theorem}

\begin{proof}
    (1)
    注意,
    \begin{align*}
        [C_{m+n}\, S(f, g; n, m)]_{1,j}
        = {} &
        \sum_{\ell = 1}^{m+n}
            {[C_{m+n}]_{1,\ell} [S(f, g; n, m)]_{\ell,j}}
        \\
        = {} &
        \sum_{1 \leq \ell \leq m+n}
        {[S(f, g; n, m)]_{\ell,j} x^{m+n-\ell}}.
    \end{align*}
    若 \(j \leq m\), 则 \([S(f, g; n, m)]_{\ell,j} = a_{\ell-j}\).
    则
    \begin{align*}
             &
        [C_{m+n}\, S(f, g; n, m)]_{1,j}
        \\
        = {} &
        \sum_{1 \leq \ell \leq m+n}
        {a_{\ell-j} x^{m+n-\ell}}
        \\
        = {} &
        \sum_{1-j \leq \ell-j \leq m+n-j}
        {a_{\ell-j} x^{n-(\ell-j)+(m-j)}}
        \\
        = {} &
        \sum_{1-j \leq h \leq n+(m-j)}
        {a_h x^{n-h}\, x^{m-j}}
        \\
        = {} &
        \sum_{0 \leq h \leq n}
        {a_h x^{n-h}\, x^{m-j}}
        +
        \sum_{\substack{
        1-j \leq h < 0 \\
                \text{或}\, n < h \leq n+(m-j)
            }}
        {a_h x^{n-h}\, x^{m-j}}
        \\
        = {} &
        \Bigg( \sum_{0 \leq h \leq n}
        {a_h x^{n-h}} \Bigg) x^{m-j}
        +
        \sum_{\substack{
        1-j \leq h < 0 \\
                \text{或}\, n < h \leq n+(m-j)
            }}
        {0\, x^{n-h}\, x^{m-j}}
        \\
        = {} &
        f(x)\, x^{m-j}.
    \end{align*}
    若 \(j > m\), 则
    \([S(f, g; n, m)]_{\ell,j} = b_{\ell-(j-m)}\).
    为方便, 记 \(k = j - m\).
    则
    \begin{align*}
             &
        [C_{m+n}\, S(f, g; n, m)]_{1,j}
        \\
        = {} &
        \sum_{1 \leq \ell \leq m+n}
        {b_{\ell-k} x^{m+n-\ell}}
        \\
        = {} &
        \sum_{1-k \leq \ell-k \leq m+n-k}
        {b_{\ell-k} x^{m-(\ell-k)+(n-k)}}
        \\
        = {} &
        \sum_{1-k \leq h \leq m+(n-k)}
        {b_h x^{m-h}\, x^{n-k}}
        \\
        = {} &
        \sum_{0 \leq h \leq m}
        {b_h x^{m-h}\, x^{n-k}}
        +
        \sum_{\substack{
        1-k \leq h < 0 \\
                \text{或}\, m < h \leq m+(n-k)
            }}
        {b_h x^{m-h}\, x^{n-k}}
        \\
        = {} &
        \Bigg( \sum_{0 \leq h \leq m}
        {b_h x^{m-h}} \Bigg) x^{n-(j-m)}
        +
        \sum_{\substack{
        1-k \leq h < 0 \\
                \text{或}\, m < h \leq m+(n-k)
            }}
        {0\, x^{m-h}\, x^{n-k}}
        \\
        = {} &
        g(x)\, x^{n-(j-m)}.
    \end{align*}

    (2)
    注意,
    \begin{align*}
             &
        [C_{m+n}\, S(f, g; n, m)\, Y]_{1,1}
        \\
        = {} &
        [(C_{m+n}\, S(f, g; n, m))\, Y]_{1,1}
        \\
        = {} &
        \sum_{j = 1}^{m+n}
            {[C_{m+n}\, S(f, g; n, m)]_{1,j} [Y]_{j,1}}
        \\
        = {} &
        \sum_{j = 1}^{m}
            {[C_{m+n}\, S(f, g; n, m)]_{1,j} [Y]_{j,1}}
        +
        \sum_{j = m+1}^{m+n}
            {[C_{m+n}\, S(f, g; n, m)]_{1,j} [Y]_{j,1}}
        \\
        = {} &
        \sum_{j = 1}^{m}
            {f(x)\, x^{m-j}\, [Y]_{j,1}}
        +
        \sum_{j = m+1}^{m+n}
            {g(x)\, x^{n-(j-m)}\, [Y]_{j,1}}
        \\
        = {} &
        f(x)
        \sum_{j = 1}^{m}
            {x^{m-j}\, [Y]_{j,1}}
        +
        g(x)
        \sum_{j = m+1}^{m+n}
            {x^{n-(j-m)}\, [Y]_{j,1}}
        \\
        = {} &
        f(x)
        \sum_{j = 1}^{m}
            {x^{m-1-(j-1)}\, [Y]_{(j-1)+1,1}}
        +
        g(x)
        \sum_{j = m+1}^{m+n}
            {x^{n-1-(j-m-1)}\, [Y]_{m+1+(j-m-1),1}}
        \\
        = {} &
        f(x)
        \sum_{k = 0}^{m-1}
            {[Y]_{k+1,1} x^{m-1-k}}
        +
        g(x)
        \sum_{k = 0}^{n-1}
            {[Y]_{m+1+k,1} x^{n-1-k}}
        \\
        = {} &
        f(x)\, u(x) + g(x)\, v(x).
    \end{align*}

    (3)
    注意,
    \begin{align*}
        [C_{m+n}\, S(f, g; n, m)\, Y]_{1,1}
        = {} &
        [C_{m+n}\, (S(f, g; n, m)\, Y)]_{1,1}
        \\
        = {} &
        \sum_{j = 1}^{m+n}
            {[C_{m+n}]_{1,j} [S(f, g; n, m)\, Y]_{j,1}}
        \\
        = {} &
        \sum_{j = 1}^{m+n}
            {x^{m+n-j} [S(f, g; n, m)\, Y]_{j,1}}
        \\
        = {} &
        \sum_{j = 1}^{m+n}
            {[S(f, g; n, m)\, Y]_{(j-1)+1,1} x^{m+n-1-(j-1)}}
        \\
        = {} &
        \sum_{k = 0}^{m+n-1}
            {[S(f, g; n, m)\, Y]_{k+1,1} x^{m+n-1-k}}.
    \end{align*}
    故, 对不超过 \(m+n-1\) 的非负整数 \(k\),
    \begin{equation*}
        [S(f, g; n, m)\, Y]_{k+1,1} = c_k.
        \qedhere
    \end{equation*}
\end{proof}

\begin{theorem}
    设 \(A\) 是 \(n\)~级阵.
    则存在 \(n \times 1\)~阵 \(P\),
    使
    \begin{align*}
        [A P]_{i,1} =
        \begin{cases}
            \det {(A)}, & i = n;       \\
            0,          & \text{别的情形}.
        \end{cases}
    \end{align*}
    进一步地, 我们可要求, \(P \neq 0\).
\end{theorem}

\begin{proof}
    设 \(W\) 是 \(A\) 的古伴 \(\operatorname{adj} {(A)}\) 的列~\(n\).
    则
    \begin{align*}
        [A W]_{i,1}
        = {} &
        \sum_{j = 1}^{n} {[A]_{i,j} [W]_{j,1}}
        \\
        = {} &
        \sum_{j = 1}^{n} {[A]_{i,j} [\operatorname{adj} {(A)}]_{j,n}}
        \\
        = {} &
        [A \operatorname{adj} {(A)}]_{i,n}
        \\
        = {} &
        [\det {(A)}\, I_n]_{i,n}
        \\
        = {} &
        \det {(A)}\, [I_n]_{i,n}.
    \end{align*}
    取 \(P\) 为 \(W\) 即可.

    设, 进一步地, 我们要求, \(P \neq 0\).
    若 \(W \neq 0\), 则我们仍取 \(P\) 为 \(W\).
    若 \(W = 0\), 则
    \(\det {(A)} = [A W]_{i,n} = [A\, 0]_{i,n} = [0]_{i,n} = 0\).
    则, 由第~1~章, 节~\malneprasekcio{23},
    或由第~1~章, 节~\malneprasekcio{25},
    或由 ``REF 阵、线性方程组、秩'',
    存在非零的 \(n \times 1\) 阵 \(Q\),
    使 \(A Q = 0\).
    取 \(P\) 为 \(Q\) 即可.
\end{proof}

\begin{theorem}
    设假定~\ref{standing-assumption} 成立.

    (1)
    存在整式 \(u\), \(v\), 使
    \begin{align*}
        f(x)\, u(x) + g(x)\, v(x) = \det {(S(f, g; n, m))}.
    \end{align*}
    注意, 上式的右侧不含 \(x\).

    (2)
    设存在数 \(c\), 使 \(f(c) = 0 = g(c)\).
    则
    \begin{align*}
        \det {(S(f, g; n, m))} = 0.
    \end{align*}
\end{theorem}

\begin{proof}
    (1)
    我们知道, 存在 \((m + n) \times 1\)~阵 \(P\),
    使
    \begin{align*}
        [S(f, g; n, m)\, P]_{i,1} =
        \begin{cases}
            \det {(S(f, g; n, m))}, & i = m + n;   \\
            0,                      & \text{别的情形}.
        \end{cases}
    \end{align*}
    记
    \begin{align*}
        u(x) = \sum_{k = 0}^{m-1} {[P]_{k+1,1} x^{m-1-k}},
        \quad
        v(x) = \sum_{k = 0}^{n-1} {[P]_{m+1+k,1} x^{n-1-k}}.
    \end{align*}
    则
    \begin{align*}
        f(x)\, u(x) + g(x)\, v(x) = \sum_{k = 0}^{m+n-1} {c_k x^{m+n-1-k}},
    \end{align*}
    其中,
    \begin{align*}
        c_k = [S(f, g; n, m)\, P]_{k+1,1}
        = \begin{cases}
              \det {(S(f, g; n, m))}, & k = m + n - 1; \\
              0,                      & \text{别的情形}.
          \end{cases}
    \end{align*}

    (2)
    我们其实已证它, 但我们现在别地证它.
    注意, 代 \(x\) 以 \(c\), 得
    \begin{equation*}
        \det {(S(f, g; n, m))}
        = f(c)\, u(c) + g(c)\, v(c)
        = 0.
        \qedhere
    \end{equation*}
\end{proof}

\begin{theorem}\label{thm:resultant-swap}
    设假定~\ref{standing-assumption} 成立.

    (1)
    \(S(g, f; m, n)\) 是 \(m + n\)~级阵,
    且 \(S(f, g; n, m)\) 的列~\(j\) 是
    \(S(g, f; m, n)\) 的列~\(n + j\)
    (若 \(j \leq m\))
    或 \(S(g, f; m, n)\) 的列~\(j - m\)
    (若 \(j > m\)).

    (2)
    \(\det {(S(g, f; m, n))}
    = (-1)^{m n} \det {(S(f, g; n, m))}\).
\end{theorem}

\begin{proof}
    注意, \(S(g, f; m, n)\) 是这样的 \(m + n\)~级阵:
    \begin{align*}
        [S(g, f; m, n)]_{i,j}
        = \begin{cases}
              b_{i-j},     & j \leq n; \\
              a_{i-(j-n)}, & j > n.
          \end{cases}
    \end{align*}
    故, 若 \(j \leq m\),
    \begin{align*}
        [S(f, g; n, m)]_{i,j}
        = a_{i-j}
        = a_{i-(n+j-n)}
        = [S(g, f; m, n)]_{i,n+j},
    \end{align*}
    且若 \(j > m\),
    \begin{align*}
        [S(f, g; n, m)]_{i,j}
        = b_{i-(j-m)}
        = [S(g, f; m, n)]_{i,j-m}.
    \end{align*}

    (2)
    若 \(m = 0\) 或 \(n = 0\),
    则这是平凡地对的
    (见例~\ref{emp:morniszero}).

    设 \(m \geq 1\) 且 \(n \geq 1\).
    对 \(S(g, f; m, n)\),
    交换列~\(n + 1\) 与列~\(n\),
    再交换新的阵的列~\(n\) 与列~\(n - 1\),
    ……
    再交换新的阵的列~\(2\) 与列~\(1\),
    我们作了 \(n\)~次相邻的列的交换,
    使 \(S(g, f; m, n)\) 的列~\(n + 1\) 到列~\(1\),
    且别的列的相对位置不变.
    类似地, 我们也可使这个阵的%
    列~\(n + 2\), \(\dots\), \(n + m\)
    分别地到%
    列~\(2\), \(\dots\), \(m\).
    则 \(S(g, f; m, n)\) 的列 \(1\), \(\dots\), \(n\)
    分别地到%
    列~\(m + 1\), \(\dots\), \(m + n\).
    则 \(n \cdot m\)~次相邻的列的交换%
    变 \(S(g, f; m, n)\) 为 \(S(f, g; n, m)\).
    则
    \(\det {(S(g, f; m, n))}
    = (-1)^{m n} \det {(S(f, g; n, m))}\).
\end{proof}

\begin{theorem}\label{thm:resultant-recursive}
    设假定~\ref{standing-assumption} 成立.
    设 \(t\) 是数, 且
    \begin{align*}
        h(x)
        = f(x)\, (x - t)
        = \sum_{k = 0}^{n+1} {c_k x^{n+1-k}}.
    \end{align*}
    我们约定, 若 \(k < 0\) 或 \(k > n+1\), 则 \(c_k = 0\).
    则
    \begin{align*}
        \det {(S(h, g; n+1, m))} = \det {(S(f, g; n, m))}\, g(t),
    \end{align*}
    且
    \begin{align*}
        \det {(S(g, h; m, n+1))} = (-1)^m \det {(S(g, f; m, n))}\, g(t).
    \end{align*}
\end{theorem}

\begin{proof}
    我们证式~1;
    您用式~1 与定理~\ref{thm:resultant-swap} 证式~2.

    注意, 由整式的乘法,
    \begin{align*}
        c_k =
        \begin{cases}
            a_0,             & k = 0;           \\
            a_k - t a_{k-1}, & 1 \leq k \leq n; \\
            -t a_n,          & k = n + 1.
        \end{cases}
    \end{align*}
    回想, 若 \(k < 0\) 或 \(k > n\),
    则 \(a_k = 0\).
    于是, 统一地, \(c_k = a_k - t a_{k-1}\).
    注意, \(a_k = c_k + t a_{k-1}\).

    我们记
    \(g_\ell (x) = 0\)
    (若 \(\ell < 0\)),
    且
    \(g_\ell (x) = g_{\ell-1} (x)\, x + b_\ell\)
    (若 \(\ell \geq 0\)).
    则对任何整数 \(\ell\),
    有 \(g_\ell (t) = b_\ell + t g_{\ell-1} (t)\).
    注意, \(b_\ell = g_\ell (t) - t g_{\ell-1} (t)\).

    注意, \(m + n + 1\) 级阵 \(S(h, g; n+1, m)\) 等于
    \begin{align*}
        \begin{bmatrix}
            c_{0}   & c_{-1}    & c_{-2}    & \cdots & c_{1-m}   & b_{0}   & b_{-1}    & \cdots & b_{-n}  \\
            c_{1}   & c_{0}     & c_{-1}    & \cdots & c_{2-m}   & b_{1}   & b_{0}     & \cdots & b_{1-n} \\
            c_{2}   & c_{1}     & c_{0}     & \cdots & c_{3-m}   & b_{2}   & b_{1}     & \cdots & b_{2-n} \\
            \vdots  & \vdots    & \vdots    & {}     & \vdots    & \vdots  & \vdots    & {}     & \vdots  \\
            c_{n}   & c_{n-1}   & c_{n-2}   & \cdots & c_{n+1-m} & b_{n}   & b_{n-1}   & \cdots & b_{0}   \\
            c_{n+1} & c_{n}     & c_{n-1}   & \cdots & c_{n+2-m} & b_{n+1} & b_{n}     & \cdots & b_{1}   \\
            \vdots  & \vdots    & \vdots    & {}     & \vdots    & \vdots  & \vdots    & {}     & \vdots  \\
            c_{m+n} & c_{m+n-1} & c_{m+n-2} & \cdots & c_{n+1}   & b_{m+n} & b_{m+n-1} & \cdots & b_{m}   \\
        \end{bmatrix},
    \end{align*}
    即
    \begin{align*}
        \begin{bmatrix}
            a_{0}   & a_{-1}    & a_{-2}    & \cdots & a_{1-m}   & g_{0} (t) & g_{-1} (t) & \cdots & g_{-n} (t) \\
            c_{1}   & c_{0}     & c_{-1}    & \cdots & c_{2-m}   & b_{1}     & b_{0}      & \cdots & b_{1-n}    \\
            c_{2}   & c_{1}     & c_{0}     & \cdots & c_{3-m}   & b_{2}     & b_{1}      & \cdots & b_{2-n}    \\
            \vdots  & \vdots    & \vdots    & {}     & \vdots    & \vdots    & \vdots     & {}     & \vdots     \\
            c_{n}   & c_{n-1}   & c_{n-2}   & \cdots & c_{n+1-m} & b_{n}     & b_{n-1}    & \cdots & b_{0}      \\
            c_{n+1} & c_{n}     & c_{n-1}   & \cdots & c_{n+2-m} & b_{n+1}   & b_{n}      & \cdots & b_{1}      \\
            \vdots  & \vdots    & \vdots    & {}     & \vdots    & \vdots    & \vdots     & {}     & \vdots     \\
            c_{m+n} & c_{m+n-1} & c_{m+n-2} & \cdots & c_{n+1}   & b_{m+n}   & b_{m+n-1}  & \cdots & b_{m}      \\
        \end{bmatrix}.
    \end{align*}
    我们加行~\(1\) 的 \(t\) 倍到行~\(2\),
    再加行~\(2\) 的 \(t\) 倍到行~\(3\),
    ……
    再加行~\(m+n\) 的 \(t\) 倍到行~\(m+n+1\),
    得 \(m+n+1\)~级阵
    \begin{align*}
        S_1 =
        \begin{bmatrix}
            a_{0}   & a_{-1}    & a_{-2}    & \cdots & a_{1-m}   & g_{0} (t)   & g_{-1} (t)    & \cdots & g_{-n} (t)  \\
            a_{1}   & a_{0}     & a_{-1}    & \cdots & a_{2-m}   & g_{1} (t)   & g_{0} (t)     & \cdots & g_{1-n} (t) \\
            a_{2}   & a_{1}     & a_{0}     & \cdots & a_{3-m}   & g_{2} (t)   & g_{1} (t)     & \cdots & g_{2-n} (t) \\
            \vdots  & \vdots    & \vdots    & {}     & \vdots    & \vdots      & \vdots        & {}     & \vdots      \\
            a_{n}   & a_{n-1}   & a_{n-2}   & \cdots & a_{n+1-m} & g_{n} (t)   & g_{n-1} (t)   & \cdots & g_{0} (t)   \\
            a_{n+1} & a_{n}     & a_{n-1}   & \cdots & a_{n+2-m} & g_{n+1} (t) & g_{n} (t)     & \cdots & g_{1} (t)   \\
            \vdots  & \vdots    & \vdots    & {}     & \vdots    & \vdots      & \vdots        & {}     & \vdots      \\
            a_{m+n} & a_{m+n-1} & a_{m+n-2} & \cdots & a_{n+1}   & g_{m+n} (t) & g_{m+n-1} (t) & \cdots & g_{m} (t)   \\
        \end{bmatrix}.
    \end{align*}
    我们知道, \(\det {(S_1)} = \det {(S(h, g; n+1, m))}\).

    我们加列~\(m+2\) 的 \(-t\) 倍到列~\(m+1\),
    再加列~\(m+3\) 的 \(-t\) 倍到列~\(m+2\),
    ……
    再加列~\(m+n+1\) 的 \(-t\) 倍到列~\(m+n\),
    得 \(m+n+1\)~级阵
    \begin{align*}
        S_2 =
        \begin{bmatrix}
            a_{0}   & a_{-1}    & a_{-2}    & \cdots & a_{1-m}   & b_{0}   & b_{-1}    & \cdots & g_{-n} (t)  \\
            a_{1}   & a_{0}     & a_{-1}    & \cdots & a_{2-m}   & b_{1}   & b_{0}     & \cdots & g_{1-n} (t) \\
            a_{2}   & a_{1}     & a_{0}     & \cdots & a_{3-m}   & b_{2}   & b_{1}     & \cdots & g_{2-n} (t) \\
            \vdots  & \vdots    & \vdots    & {}     & \vdots    & \vdots  & \vdots    & {}     & \vdots      \\
            a_{n}   & a_{n-1}   & a_{n-2}   & \cdots & a_{n+1-m} & b_{n}   & b_{n-1}   & \cdots & g_{0} (t)   \\
            a_{n+1} & a_{n}     & a_{n-1}   & \cdots & a_{n+2-m} & b_{n+1} & b_{n}     & \cdots & g_{1} (t)   \\
            \vdots  & \vdots    & \vdots    & {}     & \vdots    & \vdots  & \vdots    & {}     & \vdots      \\
            a_{m+n} & a_{m+n-1} & a_{m+n-2} & \cdots & a_{n+1}   & b_{m+n} & b_{m+n-1} & \cdots & g_{m} (t)   \\
        \end{bmatrix}.
    \end{align*}
    我们知道, \(\det {(S_2)} = \det {(S_1)}\).
    则 \(\det {(S_2)} = \det {(S(h, g; n+1, m))}\).
    注意,
    \begin{align*}
        [S_2]_{i,j} =
        \begin{cases}
            a_{i-j},       & j \leq m;         \\
            b_{i-(j-m)},   & m < j \leq m + n; \\
            g_{i-1-n} (t), & j = m + n + 1.
        \end{cases}
    \end{align*}
    于是, 当 \(i = m + n + 1\) 时,
    若 \(j \leq m\),
    则 \([S_2]_{i,j} = a_{n+1+(m-j)} = 0\)
    (注意, \(n+1+(m-j) > n\)),
    且若 \(m < j \leq m + n\),
    则 \([S_2]_{i,j} = b_{m+1+(m+n-j)} = 0\)
    (注意, \(m+1+(m+n-j) > m\)).
    注意, \(S_2 ({m+n+1}|{m+n+1}) = S(f, g; n, m)\).
    注意, \(g_m (t) = g(t)\).
    于是, 按行~\(m+n+1\) 展开, 有
    \begin{align*}
             &
        \det {(S(h, g; n+1, m))}
        \\
        = {} &
        \det {(S_2)}
        \\
        = {} &
        (-1)^{(m+n+1) + (m+n+1)} g_m (t) \det {(S_2 ({m+n+1}|{m+n+1}))}
        \\
        = {} &
        \det {(S(f, g; n, m))}\, g(t).
        \qedhere
    \end{align*}
\end{proof}

\begin{theorem}
    设假定~\ref{standing-assumption} 成立.
    若 \(f(x) = a_0 (x - x_1) \dots (x - x_n)\), 则
    \begin{align*}
        \det {(S(f, g; n, m))} = a_0^m g(x_1) g(x_2) \dots g(x_n).
    \end{align*}
    类似地,
    若 \(g(x) = b_0 (x - y_1) \dots (x - y_m)\), 则
    \begin{align*}
        \det {(S(f, g; n, m))} = (-1)^{mn} b_0^n f(y_1) f(y_2) \dots f(y_m).
    \end{align*}
\end{theorem}

\begin{proof}
    我们证式~1;
    您用式~1 与定理~\ref{thm:resultant-swap} 证式~2.

    若 \(m = 0\),
    则式~1 的左侧是 \(b_0^n\),
    且式~1 的右侧是
    \(a_0^0 g(x_1) \dots g(x_n) = 1 \cdot b_0^n\)
    (注意, \(g(x) = b_0\)).

    以下, 我们设 \(m \geq 1\).

    记 \(F_0 (x) = a_0\),
    且 \(F_i (x) = F_{i-1} (x)\, (x - x_i)\).
    则
    \begin{align*}
        \det {(S(f, g; n, m))}
        = {} &
        \det {(S(F_n, g; n, m))}
        \\
        = {} &
        \det {(S(F_{n-1}, g; n-1, m))}\,
        g(x_n)
        \\
        = {} &
        \det {(S(F_{n-2}, g; n-2, m))}\,
        g(x_{n-1}) g(x_n)
        \\
        = {} &
        \dots
        \\
        = {} &
        \det {(S(F_0, g; 0, m))}\,
        g(x_1) g(x_2) \dots g(x_n)
        \\
        = {} &
        a_0^m g(x_1) g(x_2) \dots g(x_n).
        \qedhere
    \end{align*}
\end{proof}

\begin{theorem}
    设假定~\ref{standing-assumption} 成立.

    (1)
    设存在数 \(c\), 使 \(f(c) = 0 = g(c)\).
    则 \(\det {(S(f, g; n, m))} = 0\).

    (2)
    设 \(\det {(S(f, g; n, m))} = 0\).
    则 \(a_0 = 0 = b_0\),
    或当 \(a_0 \neq 0\) 或 \(b_0 \neq 0\) 时,
    存在数 \(c\), 使 \(f(c) = 0 = g(c)\).
\end{theorem}

\begin{proof}
    (1)
    定理~\ref{thm:necessary-condition-for-the-existence-of-a-common-root}.

    (2)
    若 \(a_0 = 0 = b_0\), 则 \(\det {(S(f, g; n, m))} = 0\).

    若 \(a_0 \neq 0\), 则我们可写
    \(f(x) = a_0 (x - x_1) (x - x_2) \dots (x - x_n)\),
    其中, \(x_1\), \(x_2\), \(\dots\), \(x_n\) 是数.
    注意,
    \(f(x_1)\), \(f(x_2)\), \(\dots\), \(f(x_n)\) 都是 \(0\).
    则
    \begin{align*}
        0
        = \det {(S(f, g; n, m))}
        = a_0^m g(x_1) g(x_2) \dots g(x_n).
    \end{align*}
    因为 \(a_0 \neq 0\),
    我们说,
    在 \(g(x_1)\), \(g(x_2)\), \(\dots\), \(g(x_n)\),
    \(0\) 必存在.

    若 \(b_0 \neq 0\), 您可类似地证它:
    您用
    \begin{align*}
        0
        = \det {(S(f, g; n, m))}
        = (-1)^{m n} b_0^n f(y_1) f(y_2) \dots f(y_m),
    \end{align*}
    其中, \(g(x) = b_0 (x - y_1) (x - y_2) \dots (x - y_m)\).
\end{proof}

我们看几个例.

\begin{example}
    解方程组
    \begin{equation}
        \left\{
        \begin{aligned}
            11x^2 - 2xy - 44y^2 - x + 26y - 32  & = 0, \\
            4x^2 - 18xy + 49y^2 - 4x + 9y - 118 & = 0.
        \end{aligned}
        \right.
        \label{eq:soe1}
    \end{equation}

    设 \(x = a\), \(y = b\) 是%
    方程组~\eqref{eq:soe1} 的一个解.
    则
    \begin{align*}
         &
        \begin{aligned}
            0
            = {} & 11a^2 - 2ab - 44b^2 - a + 26b - 32             \\
            = {} & 11 a^2 + (-2 b - 1) a + (-44 b^2 + 26 b - 32),
        \end{aligned}
        \\
         &
        \begin{aligned}
            0
            = {} & 4a^2 - 18ab + 49b^2 - 4a + 9b - 118           \\
            = {} & 4 a^2 + (-18 b - 4) a + (49 b^2 + 9 b - 118).
        \end{aligned}
    \end{align*}
    我们记
    \begin{align*}
        f(x; y) & = 11 x^2 + (-2 y - 1) x + (-44 y^2 + 26 y - 32), \\
        g(x; y) & = 4 x^2 + (-18 y - 4) x + (49 y^2 + 9 y - 118).
    \end{align*}
    则 \(f(a; b) = 0 = g(a; b)\).
    故
    \begin{align*}
        \begin{bmatrix}
            11            & 0             & 4            & 0            \\
            -2b-1         & 11            & -18b-4       & 4            \\
            -44b^2+26b-32 & -2b-1         & 49b^2+9b-118 & -18b-4       \\
            0             & -44b^2+26b-32 & 0            & 49b^2+9b-118 \\
        \end{bmatrix}
    \end{align*}
    的行列式是 \(0\).
    可以算出, 此阵的行列式是
    \(342\,125\, (b^2 - 1)\, (b^2 - 4)\).
    于是, 若 \(x = a\), \(y = b\) 是%
    方程组~\eqref{eq:soe1} 的一个解,
    则 \(b = 1\) 或 \(b = -1\)
    或 \(b = 2\) 或 \(b = -2\).
    则 (代 \(b\) 以 \(1\))
    \begin{equation}
        \left\{
        \begin{aligned}
            11a^2 - 2a - 44 - a + 26 - 32  & = 0, \\
            4a^2 - 18a + 49 - 4a + 9 - 118 & = 0;
        \end{aligned}
        \right.
        \label{eq:soe2}
    \end{equation}
    或 (代 \(b\) 以 \(-1\))
    \begin{equation}
        \left\{
        \begin{aligned}
            11a^2 + 2a - 44 - a - 26 - 32  & = 0, \\
            4a^2 + 18a + 49 - 4a - 9 - 118 & = 0;
        \end{aligned}
        \right.
        \label{eq:soe3}
    \end{equation}
    或 (代 \(b\) 以 \(2\))
    \begin{equation}
        \left\{
        \begin{aligned}
            11a^2 - 4a - 176 - a + 52 - 32   & = 0, \\
            4a^2 - 36a + 196 - 4a + 18 - 118 & = 0;
        \end{aligned}
        \right.
        \label{eq:soe4}
    \end{equation}
    或 (代 \(b\) 以 \(-2\))
    \begin{equation}
        \left\{
        \begin{aligned}
            11a^2 + 4a - 176 - a - 52 - 32   & = 0, \\
            4a^2 + 36a + 196 - 4a - 18 - 118 & = 0.
        \end{aligned}
        \right.
        \label{eq:soe5}
    \end{equation}
    方程组~\eqref{eq:soe2} 的解是 \(a = -2\);
    方程组~\eqref{eq:soe3} 的解是 \(a = 3\);
    方程组~\eqref{eq:soe4} 的解是 \(a = 4\);
    方程组~\eqref{eq:soe5} 的解是 \(a = -5\).
    于是, 若 \(x = a\), \(y = b\) 是%
    方程组~\eqref{eq:soe1} 的一个解,
    必:
    \begin{equation*}
        \begin{alignedat}{7}
            {}             & a & {} = {} & {-2}, & \quad & b & {} = {} & 1;    \\
            \text{或} \quad & a & {} = {} & 3,    & \quad & b & {} = {} & {-1}; \\
            \text{或} \quad & a & {} = {} & 4,    & \quad & b & {} = {} & 2;    \\
            \text{或} \quad & a & {} = {} & {-5}, & \quad & b & {} = {} & {-2}.
        \end{alignedat}
    \end{equation*}
    最后, 我们验证这些是否是%
    方程组~\eqref{eq:soe1} 的解.
    可以验证, 它们都是解.
    于是,
    方程组~\eqref{eq:soe1} 的解是:
    \begin{equation*}
        \begin{alignedat}{7}
            {}             & x & {} = {} & {-2}, & \quad & y & {} = {} & 1;    \\
            \text{或} \quad & x & {} = {} & 3,    & \quad & y & {} = {} & {-1}; \\
            \text{或} \quad & x & {} = {} & 4,    & \quad & y & {} = {} & 2;    \\
            \text{或} \quad & x & {} = {} & {-5}, & \quad & y & {} = {} & {-2}.
        \end{alignedat}
    \end{equation*}
\end{example}

\begin{example}
    设数 \(X\), \(Y\), \(T\) 适合
    \begin{equation}
        X = \frac{1 + 8T - T^2}{1 + T^2},
        \quad
        Y = \frac{4T}{1 + T^2}.
        \label{eq:soe6}
    \end{equation}
    我们可消去 \(T\), 以得到 \(X\) 与 \(Y\) 的关系:
    若 \((X, Y, T) = (x, y, t)\) 适合式~\eqref{eq:soe6},
    则 \((X, Y) = (x, y)\) 适合某关系 (不含 \(T\)),
    且反过来,
    若 \((X, Y) = (x, y)\) 适合这个不含 \(T\) 的关系,
    则存在 \(t\),
    使 \((X, Y, T) = (x, y, t)\) 适合式~\eqref{eq:soe6}.

    设 \((X, Y, T) = (x, y, t)\) 适合式~\eqref{eq:soe6}.
    去分母, 有
    \begin{align*}
        (1 + t^2) x - (1 + 8t - t^2) & = 0, \\
        (1 + t^2) y - 4t             & = 0,
    \end{align*}
    即
    \begin{align*}
        (x + 1) t^2 + (-8) t + (x - 1) & = 0, \\
        y t^2 + (-4) t + y             & = 0.
    \end{align*}
    记
    \begin{align*}
        f(T; X, Y) & = (X + 1) T^2 + (-8) T + (X - 1), \\
        g(T; X, Y) & = Y T^2 + (-4) T + Y.
    \end{align*}
    则 \(f(t; x, y) = 0 = g(t; x, y)\).
    反过来,
    若 \((X, Y, T) = (x, y, t)\) 适合
    \(f(T; X, Y) = 0 = g(T; X, Y)\),
    则 \(t^2 + 1 \neq 0\).
    (反证法: 若 \(y t^2 - 4 t + y = 0\),
    且 \(t^2 + 1 = 0\),
    则 \(4 t = 0\).
    则 \(t = 0\).
    可是, \(0^2 + 1 \neq 0\).)
    则 \((X, Y, T) = (x, y, t)\) 适合式~\eqref{eq:soe6}.

    所以,
    若 \((X, Y, T) = (x, y, t)\) 适合式~\eqref{eq:soe6},
    则
    \begin{align*}
        0 =
        \det {
            \begin{bmatrix}
                x+1 & 0   & y  & 0  \\
                -8  & x+1 & -4 & y  \\
                x-1 & -8  & y  & -4 \\
                0   & x-1 & 0  & y  \\
            \end{bmatrix}
        }
        = 4 (4 x^2 - 16 x y + 17 y^2 - 4).
    \end{align*}
    记
    \begin{align*}
        R(X, Y) = 4 X^2 - 16 X Y + 17 Y^2 - 4.
    \end{align*}
    则 \(R(x, y) = 0\).
    反过来,
    若 \((X, Y) = (x, y)\) 适合 \(R(X, Y) = 0\),
    则 \(x + 1 = 0 = y\),
    或 \(x + 1 \neq 0\) 或 \(y \neq 0\) 时,
    存在数 \(t\),
    使 \(f(t; x, y) = 0 = g(t; x, y)\).
    所以,
    若 \((X, Y) = (x, y)\) 适合 \(R(X, Y) = 0\),
    则 \((x, y) = (-1, 0)\),
    或 \((x, y) \neq (-1, 0)\) 时,
    存在数 \(t\),
    使 \((X, Y, T) = (x, y, t)\) 适合式~\eqref{eq:soe6}.
    我们可以验证, 不存在数 \(t\),
    使 \((X, Y, T) = (-1, 0, t)\) 适合式~\eqref{eq:soe6}.
    (反证法: 若 \(0 = 4t/(1 + t^2)\),
    则 \(t = 0\).
    可是, \(-1 \neq (1 + 8 \cdot 0 - 0^2) / (1 + 0^2)\).)

    综上, \(X\), \(Y\) 的关系是
    \begin{align*}
        4 X^2 - 16 X Y + 17 Y^2 - 4 = 0,
        \quad
        (X, Y) \neq (-1, 0).
    \end{align*}
\end{example}

\begin{theorem}
    设假定~\ref{standing-assumption} 成立.
    设 \(m \geq n\).

    设
    \begin{align*}
        q(x) = \sum_{k = 0}^{m-n} {c_k x^{m-n-k}}.
    \end{align*}
    为方便, 我们约定,
    若 \(k < 0\) 或 \(k > m - n\), 则 \(c_k = 0\).

    设
    \begin{align*}
        p(x) = g(x) + f(x) q(x) = \sum_{k = 0}^{m} {d_k x^{m-k}}.
    \end{align*}
    为方便, 我们约定,
    若 \(k < 0\) 或 \(k > m\), 则 \(d_k = 0\).

    则
    \begin{align*}
         & \det {(S(f, p; n, m))} = \det {(S(f, g; n, m))}, \\
         & \det {(S(p, f; m, n))} = \det {(S(g, f; m, n))}.
    \end{align*}
\end{theorem}

\begin{proof}
    我们证式~1;
    您用式~1 与定理~\ref{thm:resultant-swap} 证式~2.

    若 \(n = 0\),
    则 \(S(f, p; 0, m)\) 与 \(S(f, g; 0, m)\)
    都是 \(a_0 I_m\).
    则它们的行列式相等.

    以下, 我们设 \(n \geq 1\).

    由整式的加法与乘法,
    \begin{align*}
        d_k =
        \begin{dcases}
            b_k + \sum_{\ell = 0}^{k} {a_{k-\ell} c_{\ell}},
             & 0 \leq k \leq m; \\
            0,
             & \text{别的情形}.
        \end{dcases}
    \end{align*}
    注意, 若 \(0 \leq k \leq m\),
    \begin{align*}
        \sum_{\ell = 0}^{k} {a_{k-\ell} c_{\ell}}
        = \sum_{\ell = 0}^{m-n} {a_{k-\ell} c_{\ell}}.
    \end{align*}
    (若 \(k \leq m-n\),
    则当 \(\ell > k\) 时, \(a_{k-\ell} = 0\).
    若 \(k > m-n\),
    则当 \(\ell > m - n\) 时, \(c_{\ell} = 0\)).
    并且, 当 \(k < 0\) 或 \(k > m\) 时,
    \begin{align*}
        \sum_{\ell = 0}^{m-n} {a_{k-\ell} c_{\ell}} = 0.
    \end{align*}
    (若 \(k < 0\),
    则当 \(\ell \geq 0\) 时, \(k - \ell < 0\),
    故 \(a_{k-\ell} = 0\).
    若 \(k > m\),
    则当 \(\ell \leq m - n\) 时,
    \(k - \ell > m - \ell \geq m - (m - n) = n\),
    故 \(a_{k-\ell} = 0\).)
    于是, 统一地, 我们可写
    \begin{align*}
        d_k = b_k + \sum_{\ell = 0}^{m-n} {a_{k-\ell} c_{\ell}}.
    \end{align*}
    我们考虑 \(\det {(S(f, p; n, m))}\)
    与 \(\det {(S(f, g; n, m))}\)
    的关系.

    注意, \(m + n\) 级阵 \(S(f, g; n, m)\) 等于
    \begin{align*}
        \begin{bmatrix}
            a_{0}     & a_{-1}    & a_{-2}    & \cdots & a_{1-m}   & b_{0}     & b_{-1}    & \cdots & b_{1-n} \\
            a_{1}     & a_{0}     & a_{-1}    & \cdots & a_{2-m}   & b_{1}     & b_{0}     & \cdots & b_{2-n} \\
            a_{2}     & a_{1}     & a_{0}     & \cdots & a_{3-m}   & b_{2}     & b_{1}     & \cdots & b_{3-n} \\
            \vdots    & \vdots    & \vdots    & {}     & \vdots    & \vdots    & \vdots    & {}     & \vdots  \\
            a_{n-1}   & a_{n-2}   & a_{n-3}   & \cdots & a_{n-m}   & b_{n-1}   & b_{n-2}   & \cdots & b_{0}   \\
            a_{n}     & a_{n-1}   & a_{n-2}   & \cdots & a_{n+1-m} & b_{n}     & b_{n-1}   & \cdots & b_{1}   \\
            a_{n+1}   & a_{n}     & a_{n-1}   & \cdots & a_{n+2-m} & b_{n+1}   & b_{n}     & \cdots & b_{2}   \\
            \vdots    & \vdots    & \vdots    & {}     & \vdots    & \vdots    & \vdots    & {}     & \vdots  \\
            a_{m+n-1} & a_{m+n-2} & a_{m+n-3} & \cdots & a_{n}     & b_{m+n-1} & b_{m+n-2} & \cdots & b_{m}   \\
        \end{bmatrix},
    \end{align*}
    且 \(m + n\) 级阵 \(S(f, p; n, m)\) 等于
    \begin{align*}
        \begin{bmatrix}
            a_{0}     & a_{-1}    & a_{-2}    & \cdots & a_{1-m}   & d_{0}     & d_{-1}    & \cdots & d_{1-n} \\
            a_{1}     & a_{0}     & a_{-1}    & \cdots & a_{2-m}   & d_{1}     & d_{0}     & \cdots & d_{2-n} \\
            a_{2}     & a_{1}     & a_{0}     & \cdots & a_{3-m}   & d_{2}     & d_{1}     & \cdots & d_{3-n} \\
            \vdots    & \vdots    & \vdots    & {}     & \vdots    & \vdots    & \vdots    & {}     & \vdots  \\
            a_{n-1}   & a_{n-2}   & a_{n-3}   & \cdots & a_{n-m}   & d_{n-1}   & d_{n-2}   & \cdots & d_{0}   \\
            a_{n}     & a_{n-1}   & a_{n-2}   & \cdots & a_{n+1-m} & d_{n}     & d_{n-1}   & \cdots & d_{1}   \\
            a_{n+1}   & a_{n}     & a_{n-1}   & \cdots & a_{n+2-m} & d_{n+1}   & d_{n}     & \cdots & d_{2}   \\
            \vdots    & \vdots    & \vdots    & {}     & \vdots    & \vdots    & \vdots    & {}     & \vdots  \\
            a_{m+n-1} & a_{m+n-2} & a_{m+n-3} & \cdots & a_{n}     & d_{m+n-1} & d_{m+n-2} & \cdots & d_{m}   \\
        \end{bmatrix}.
    \end{align*}
    注意, \(j > m\) 时,
    \begin{align*}
        d_{i-(j-m)}
        = {} &
        b_{i-(j-m)} + \sum_{0 \leq \ell \leq m-n} {a_{i-(j-m)-\ell} c_{\ell}}
        \\
        = {} &
        b_{i-(j-m)} + \sum_{j-m \leq j-m+\ell \leq j-n} {a_{i-(j-m+\ell)} c_{(j-m+\ell)-(j-m)}}
        \\
        = {} &
        b_{i-(j-m)} + \sum_{j-m \leq r \leq j-m+(m-n)} {c_{r-(j-m)} a_{i-r}}.
    \end{align*}
    于是, 对 \(S(f, g; n, m)\), 我们%
    加列~\(1\) 的 \(c_0\)~倍到列~\(m+1\),
    再加列~\(2\) 的 \(c_1\)~倍到列~\(m+1\),
    ……
    再加列~\(m-n+1\) 的 \(c_{m-n}\)~倍到列~\(m+1\),
    且我们加列~\(2\) 的 \(c_0\)~倍到列~\(m+2\),
    再加列~\(3\) 的 \(c_1\)~倍到列~\(m+2\),
    ……
    再加列~\(m-n+2\) 的 \(c_{m-n}\)~倍到列~\(m+2\),
    ……
    且我们加列~\(n\) 的 \(c_0\)~倍到列~\(m+n\),
    再加列~\(n+1\) 的 \(c_1\)~倍到列~\(m+n\),
    ……
    再加列~\(m\) 的 \(c_{m-n}\)~倍到列~\(m+n\),
    则我们得 \(S(f, p; n, m)\).
    则
    \begin{equation*}
        \det {(S(f, p; n, m))} = \det {(S(f, g; n, m))}.
        \qedhere
    \end{equation*}
\end{proof}

\begin{example}
    设假定~\ref{standing-assumption} 成立.
    设
    \begin{align*}
        G(x) = \sum_{k = 0}^{m+1} {c_k x^{m+1-k}},
    \end{align*}
    其中, \(c_k = b_{k-1}\);
    特别地, \(c_0 = 0\).
    注意, 其实, \(G(x) = g(x)\),
    但
    \begin{align*}
        \sum_{k = 0}^{m} {b_k x^{m-k}}
        \quad \text{与} \quad
        \sum_{k = 0}^{m+1} {c_k x^{m+1-k}}
    \end{align*}
    的形式是不同的.
    \(S(f, g; n, m)\) 与 \(S(f, G; n, m+1)\)
    分别是 \(m+n\)~级与 \(m+n+1\)~级阵.
    我们考虑 \(\det {(S(f, g; n, m))}\)
    与 \(\det {(S(f, G; n, m+1))}\)
    的关系.

    首先, 既然 \(c_0 = 0\),
    则 \(k \leq 0\) 时, \(c_k \leq 0\).
    则 \(j > 1\) 时, \([S(f, G; n, m+1)]_{1,j} = 0\).
    (若 \(2 \leq j \leq m+1\),
    则 \([S(f, G; n, m+1)]_{1,j} = a_{1-j} = 0\);
    若 \(j \geq m+2\), 则 \([S(f, G; n, m+1)]_{1,j}
    = c_{1-(j-(m+1))} = c_{m+2-j} = 0\).)

    另一方面,
    \((S(f, G; n, m+1))(1|1) = S(f, g; n, m)\).
    注意,
    \begin{align*}
        [S(f, G; n, m+1)]_{i+1,j+1}
        = {} &
        \begin{cases}
            a_{(i+1)-(j+1)},       & j+1 \leq m+1; \\
            c_{(i+1)-(j+1-(m+1))}, & j+1 > m+1
        \end{cases}
        \\
        = {} &
        \begin{cases}
            a_{i-j},       & j \leq m; \\
            c_{i-(j-m)+1}, & j > m
        \end{cases}
        \\
        = {} &
        \begin{cases}
            a_{i-j},     & j \leq m; \\
            b_{i-(j-m)}, & j > m
        \end{cases}
        \\
        = {} &
        [S(f, g; n, m)]_{i,j}.
    \end{align*}
    所以, 由按行~\(1\) 展开,
    \begin{align*}
             &
        \det {(S(f, G; n, m+1))}
        \\
        = {} &
        (-1)^{1+1}\, [S(f, G; n, m+1)]_{1,1}
        \det {((S(f, G; n, m+1))(1|1))}
        \\
        = {} &
        a_0 \det {(S(f, g; n, m))}.
    \end{align*}

    注意, \(g = G\).
    为方便, 我们直接地写
    \begin{align*}
        \det {(S(f, g; n, m+1))} = a_0 \det {(S(f, g; n, m))}.
    \end{align*}
    由此, 若 \(d\) 是非负整数,
    \begin{align*}
        \det {(S(f, g; n, m+d))}
        = {} & a_0 \det {(S(f, g; n, m+d-1))}   \\
        = {} & a_0^2 \det {(S(f, g; n, m+d-2))} \\
        = {} & \dots                            \\
        = {} & a_0^d \det {(S(f, g; n, m))},
    \end{align*}
    其中, \(S(f, g; n, m+d)\) 的 \(g\) 被认为是
    \begin{align*}
        g(x) = \sum_{k = 0}^{m+d} {w_k x^{m+d-k}},
    \end{align*}
    其中, \(w_k = b_{k-d}\).

    最后, 类似地,
    \begin{align*}
        \det {(S(f, g; n+d, m))}
        = {} &
        (-1)^{m (n+d)} \det {(S(g, f; m, n+d))}
        \\
        = {} &
        (-1)^{m n} (-1)^{m d} b_0^d \det {(S(g, f; m, n))}
        \\
        = {} &
        (-1)^{m d} b_0^d \det {(S(f, g; n, m))}.
    \end{align*}
\end{example}

\begin{example}
    设假定~\ref{standing-assumption} 成立.
    设
    \begin{align*}
        F(x) = \sum_{k = 0}^{n} {c_k x^{n-k}},
        \quad
        G(x) = \sum_{k = 0}^{m} {d_k x^{m-k}},
    \end{align*}
    其中, (对任何整数 \(\ell\),)
    \(c_\ell = a_{n-\ell}\),
    且 \(d_\ell = b_{m-\ell}\).
    于是,
    \begin{align*}
        F(x) & = a_n x^n + a_{n-1} x^{n-1} + \dots + a_0, \\
        G(x) & = b_m x^m + b_{m-1} x^{m-1} + \dots + b_0.
    \end{align*}
    我们考虑
    \(\det {(S(F, G; n, m))}\)
    与
    \(\det {(S(f, g; n, m))}\)
    的关系.

    注意, 若 \(j \leq m\), 则
    \begin{align*}
        [S(F, G; n, m)]_{i,j}
        = {} & c_{i-j}                          \\
        = {} & a_{n-(i-j)}                      \\
        = {} & a_{n-i+j}                        \\
        = {} & a_{m+n+1-i+j-m-1}                \\
        = {} & a_{(m+n+1-i)-(m+1-j)}            \\
        = {} & [S(f, g; n, m)]_{m+n+1-i,m+1-j};
    \end{align*}
    若 \(j > m\), 则
    \begin{align*}
        [S(F, G; n, m)]_{i,j}
        = {} & d_{i-(j-m)}                            \\
        = {} & b_{m-(i-(j-m))}                        \\
        = {} & b_{m-i+(j-m)}                          \\
        = {} & b_{m+n+1-i+(j-m)-n-1}                  \\
        = {} & b_{(m+n+1-i)-(n+1-(j-m))}              \\
        = {} & b_{(m+n+1-i)-(m+n+1-(j-m)-m)}          \\
        = {} & [S(f, g; n, m)]_{m+n+1-i,m+n+1-(j-m)}.
    \end{align*}
    作 \(m+n\)~级阵 \(T\), 使
    \([T]_{i,j} = [S(f, g; n, m)]_{m+n+1-i,j}\).
    则 \(T\) 的行~\(i\) 是 \(S(f, g; n, m)\) 的行~\(m+n+1-i\).
    注意,
    \begin{align*}
        [S(F, G; n, m)]_{i,j} =
        \begin{cases}
            [T]_{i,m+1-j},       & j \leq m; \\
            [T]_{i,m+n+1-(j-m)}, & j > m.
        \end{cases}
    \end{align*}
    则 \(T\) 的列~\(j\) 是 \(S(F, G; n, m)\) 的列~\(m+1-j\)
    (若 \(j \leq m\))
    或列~\(m+n+1-(j-m)\)
    (若 \(j > m\)).
    我们分别考虑
    \(\det {(T)}\), \(\det {(S(f, g; n, m))}\) 的关系,
    与
    \(\det {(T)}\), \(\det {(S(F, G; n, m))}\)
    的关系.

    对 \(S(f, g; n, m)\),
    我们交换行~\(m+n\), \(m+n-1\),
    再交换行~\(m+n-1\), \(m+n-2\),
    ……
    再交换行~\(2\), \(1\),
    得 \(m+n\)~级阵 \(T_1\).
    则 \(T_1\) 的行~\(1\) 是 \(S(f, g; n, m)\) 的行~\(m+n\),
    且 \(T_1\) 的行~\(\ell\) 是 \(S(f, g; n, m)\) 的行~\(\ell-1\)
    (若 \(\ell > 1\)).
    我们作了 \(m+n-1\) 次相邻的行的交换.
    则 \(\det {(T_1)} = (-1)^{m+n-1} \det {(S(f, g; n, m))}\).
    注意, \(T_1\) 的行~\(1\) 是 \(T\) 的行~\(1\).

    我们可类似地使 \(T_1\)~的行~\(m+n\)
    (其是 \(S(f, g; n, m)\) 的行~\(m+n-1\))
    到行~\(2\) 的位置,
    使 \(T_1\) 的行~\(1\) 仍在行~\(1\) 的位置,
    且使 \(T_1\) 的行~\(\ell\) 在行~\(\ell+1\) 的位置
    (若 \(2 < \ell < m+n\)),
    得 \(m+n\)~级阵 \(T_2\).
    我们作了 \(m+n-2\) 次相邻的行的交换.
    则
    \begin{align*}
        \det {(T_2)}
        = {} & (-1)^{m+n-2} \det {(T_1)}                      \\
        = {} & (-1)^{(m+n-1)+(m+n-2)} \det {(S(f, g; n, m))}.
    \end{align*}

    ……

    最后, 我们共作
    \begin{align*}
        (m+n-1) + (m+n-2) + \dots + 1
        = {} & \frac{(m+n)(m+n-1)}{2}                   \\
        = {} & \frac{m(m-1)}{2} + \frac{n(n-1)}{2} + mn
    \end{align*}
    次相邻的行的交换,
    使 \(S(f, g; n, m)\) 的行~\(\ell\)
    到行~\(m+n-1-\ell\) 的位置,
    \(\ell = 1\), \(2\), \(\dots\), \(m+n\),
    得 \(m+n\)~级阵 \(T\).
    则
    \begin{align*}
        \det {(T)}
        = (-1)^{m(m-1)/2} (-1)^{n(n-1)/2} (-1)^{mn}
        \det {(S(f, g; n, m))}.
    \end{align*}

    然后, 我们可类似地,
    作 \(m + (m-1) + \dots + 1 = m(m-1)/2\)
    次相邻的列的交换,
    且再作 \(n + (n-1) + \dots + 1 = n(n-1)/2\)
    次相邻的列的交换,
    使 \(T\) 的列~\(1\), \(2\), \(\dots\), \(m\)
    分别地到列~\(m\), \(m-1\), \(\dots\), \(1\) 的位置,
    且使 \(T\) 的列~\(m+1\), \(m+2\), \(\dots\), \(m+n\)
    分别地到列~\(m+n\), \(m+n-1\), \(\dots\), \(m+1\) 的位置,
    得 \(m+n\)~级阵 \(S(F, G; n, m)\).
    则
    \begin{align*}
             &
        \det {(S(F, G; n, m))}
        \\
        = {} &
        (-1)^{n(n-1)/2} (-1)^{m(m-1)/2} \det {(T)}
        \\
        = {} &
        (-1)^{n(n-1)/2} (-1)^{m(m-1)/2}
        (-1)^{m(m-1)/2} (-1)^{n(n-1)/2} (-1)^{mn}
        \det {(S(f, g; n, m))}
        \\
        = {} &
        (-1)^{mn} \det {(S(f, g; n, m))}.
    \end{align*}

    我们以 \(a_\ell\), \(b_\ell\)
    直接地表示 \(S(F, G; n, m)\) 的元:
    \begin{align*}
        [S(F, G; n, m)]_{i,j}
        = {} &
        \begin{cases}
            a_{n-(i-j)},     & j \leq m; \\
            b_{m-(i-(j-m))}, & j > m;
        \end{cases}
        \\
        = {} &
        \begin{cases}
            a_{j-i+n}, & j \leq m; \\
            b_{j-i},   & j > m.
        \end{cases}
    \end{align*}
    对适合假定~\ref{standing-assumption} 的 \(f\), \(g\),
    定义 \(m + n\)~级阵 \(S'(f, g; n, m)\),
    其中,
    \begin{align*}
        [S'(f, g; n, m)]_{i,j}
        = \begin{cases}
              a_{j-i+n}, & j \leq m; \\
              b_{j-i},   & j > m.
          \end{cases}
    \end{align*}
    比如, 若 \(n = 2\), 且 \(m = 5\),
    则 \(S'(f, g; 2, 5)\) 等于
    \begin{align*}
        \begin{bmatrix}
            a_2    & a_3    & a_4    & a_5    & a_6 & b_5    & b_6 \\
            a_1    & a_2    & a_5    & a_4    & a_5 & b_4    & b_5 \\
            a_0    & a_1    & a_2    & a_3    & a_4 & b_3    & b_4 \\
            a_{-1} & a_0    & a_1    & a_2    & a_3 & b_2    & b_3 \\
            a_{-2} & a_{-1} & a_0    & a_1    & a_2 & b_1    & b_2 \\
            a_{-3} & a_{-2} & a_{-1} & a_0    & a_1 & b_0    & b_1 \\
            a_{-4} & a_{-3} & a_{-2} & a_{-1} & a_0 & b_{-1} & b_0 \\
        \end{bmatrix}
        =
        \begin{bmatrix}
            a_2 & 0   & 0   & 0   & 0   & b_5 & 0   \\
            a_1 & a_2 & 0   & 0   & 0   & b_4 & b_5 \\
            a_0 & a_1 & a_2 & 0   & 0   & b_3 & b_4 \\
            0   & a_0 & a_1 & a_2 & 0   & b_2 & b_3 \\
            0   & 0   & a_0 & a_1 & a_2 & b_1 & b_2 \\
            0   & 0   & 0   & a_0 & a_1 & b_0 & b_1 \\
            0   & 0   & 0   & 0   & a_0 & 0   & b_0 \\
        \end{bmatrix}.
    \end{align*}

    在一些文献, 一些作者的
    ``\(f\), \(g\) 的 Sylvester 阵'' 是这儿的
    \(S'(f, g; n, m)\),
    且 ``\(f\), \(g\) 的结式'' 是
    \(\det {(S'(f, g; n, m))}\).
    % 形式地, \(S(f, g; n, m)\) 是 ``降幂的''
    % (\(a_0\), \(a_1\), \(\dots\), \(a_n\)
    % 分别是 \(f(x)\) 的
    % \(x^n\), \(x^{n-1}\), \(\dots\), \(1\) 项的系数,
    % 且 \(b_0\), \(b_1\), \(\dots\), \(b_m\)
    % 分别是 \(g(x)\) 的
    % \(x^m\), \(x^{m-1}\), \(\dots\), \(1\) 项的系数),
    % 而 \(S'(f, g; n, m)\) 是 ``升幂的''
    % (\(a_n\), \(a_{n-1}\), \(\dots\), \(a_0\)
    % 分别是 \(f(x)\) 的
    % \(1\), \(x\), \(\dots\), \(x^n\) 项的系数,
    % 且 \(b_m\), \(b_{m-1}\), \(\dots\), \(b_0\)
    % 分别是 \(g(x)\) 的
    % \(1\), \(x\), \(\dots\), \(x^m\) 项的系数).
    % 由前面的讨论, 它们的行列式
    % (即 \(f\), \(g\) 的二类结式)
    由前面的讨论, 它们的行列式%
    至多差正负号:
    \(\det {(S'(f, g; n, m))}
    = (-1)^{m n} \det {(S(f, g; n, m))}\).
\end{example}

若您想了解结式的更多的性质, 您可以找相关的文献.
我就说这么多.

\end{document}

This material discusses
the resultant and the matrix of Sylvester.
